A MATH 5 Practice Test 4 NAME: SOLUTIONS CRN: Multiple Choice No partial credit will be given. Clearly circle one answer. No calculator!. Which of the following must be true (you may select more than one answer choice). (A) d = ln (B) A function F is called an antiderivative of f on an interval if F () = f () for all in that interval. (C) If G() = t sin(t) dt, then G() =. (D) Let f be a continuous function on [, 4]. If f() 5, then 4 f() d > 6. C. Consider the following equation involving f and its first derivative f : f () + f() =, for all real numbers. Which of the following must be true of such a function f? (A) f() =. (B) f() = e. (C) f is an integrable function. (D) There is no such function f. C. The graph of y = f() of the function f with domain (, ) is given below. 4 y y = f() 4 6 8 4 Which of the following is the area of the region between the graph of f and the -ais between = and = 9? (A) 4 (B) (C) 8 (D) 6 B
Free Response Show reasoning that is complete and correct by the standards of this course. Whenever using theorems, you should eplicitly check that all hypotheses are satisfied. Improper use of (or the absence of) proper notation will be penalized. No calculator! 4. An eperimental and theoretical physicist are studying a particle moving along a fied line. (a) Taking measurements of the particle s velocity (in ft/sec), the eperimentalist obtains the following data: t (sec) v(t) (ft/sec).5 5.5. 45.5 6. Approimate the particle s position after minute, provided the particle s initial position is s =. The position after minute (= 6 seconds) is s(6) = We can approimate the integral with a Riemann sum. Method (using a left-hand sum): Method (using a right-hand sum): 6 6 6 v(t) dt v(t) dt 5(.5 +.5 +. +.5) = 5(.9) =.5 v(t) dt 5(.5 +. +.5 +.) = 5(.49) = 4. +. = 6. (b) Using general principles, the theoretical physicist can write the position of the particle as an integral equation: t/5 s(t) = 4 + d Compute the velocity of the particle at time t = minute. Does the predicted velocity agree with the eperimentally determined velocity? Since v(t) = ds/dt, we can use the Fundamental Theorem of Calculus to obtain ( t/5 ) d dt [s] = ( t/5 ) 4 d [ ] t/5 dt + ( ) = ( ) (t/5) + 5 Therefore, v(6) = ( ) (6/5) + So the eperimentally obtained, and the predicted velocity agree. ( ) = 5 (7)(5).
(c) Using the integral equation from part (b), compute the position of the particle at time t = minute. Does the predicted position agree with your approimation from part (a)? (Note: tan ().7 and tan (4).6.) Since 6/5 s(6) = we can use u =, du = as a u-substitution: 4 + d = ( ) + d = 4 u + du = tan (u) ] 4 ( ) + d = tan (4).66 This should not agree with any reasonable approimation based on the given data; since v(t) is decreasing, the left-hand sum is an over-estimate and the right-hand sum is an underestimate. However,.66 is outside of the range given by our approimations (6. s(6).5).
5. Compute each of the following integrals: u (a) u u du u u u du = u u (u ) du = u du = u + C (b) ( ) d (c) ( ) d = = tan() + csc() [csc() + 5 cot()] + d ( ) d + 4 d + ( ) d d = ( ) ] + ( ) ] = + = tan() + csc() [csc() + 5 cot()] + d = sin() cos() d + csc () d + 5 csc() cot() d + d = ln cos() cot() 5 csc() + ln + C (d) π/4 cos() d sin () Since cos() > on the interval [, π/4], we have cos () = cos(). Thus, cos() = cos() sin () cos () = Therefore, π/4 cos() d = sin () π/4 d = π 4 4
6. Suppose f is continuous, f() =, f() =, f () > for all in the interval (, ), and f() d = /. (a) Sketch the graph of a function f satisfying the properties above. One such function is f() =. Note, f() =, f() =, f () = > for all >, and f() d = d = ] = The sketch of any function which has the interpolation properties (f() = and f() = ), is increasing, and concave up over part of the subinterval (otherwise the integral would be at least /) would be a reasonable answer. (b) What is the graphical relationship (on your sketch from the previous part) between y = f() and = f (y)? They have the same graph in the plane. y = f() has as the independent variable, but = f (y) has y as the independent variable. (c) Find the value of the integral f (y) dy. By part (b), the integral f (y) dy is complementary to f() d within the unit square [, ] [, ] (since they compute the areas of complementary regions in the square). That is, f (y) dy = f() d = = You intuition may be aided by a picture: refer to the picture you drew in part (a) and your answer from part (b). 7. Suppose that f and g are integrable and that 4 f() d = 6, 4 f() d =, 4 g() d =. Find the values of the following definite integrals, or indicate that there is not enough information to do so. (a) (b) (c) 4 f() d ( ) ( 4 f() d = f() d = f() d g() d g() d = because this is a zero width integral. f() g() d There is not enough information to determine the value of this integral. 4 ) f() d = (6 ) = ( 4) = 5